# Top 10 Current Divider Rule MCQ Problems

Basic Electrical Engineering MCQ Questions on Current divider rule.

## SI unit of current is

SI unit of current is:

- volts
- ohms
- watts
- amps

Correct answer: 4. amps

## Current always divide in

Current always divides in:

- Series
- Parallel

Correct answer: 2. Parallel

## Current divider rule is applicable on

Current divider rule is applicable on:

- Series circuits
- Parallel circuits

Correct answer: 2. Parallel

## For two parallel resistors having same resistance the current will

For two parallel resistors having same resistance the current will:

- Remain same
- Divide

Correct answer: 1. Remain same

## Which of the following is correct formula for current divider

Which of the following is correct formula for current divider:

- I
_{x}= (R_{x}/R_{eq}) * I_{t} - I
_{x}= (R_{eq}/R_{x}) * I_{t} - Both of these
- None of these

Correct answer: 2. I_{x} = (R_{eq}/R_{x}) * I_{t}

## Current Divider Rule MCQ#6

For current divider circuit give below, if R_{1} = R_{2} then:

- I
_{1}> I_{2} - I
_{1}< I_{2} - I
_{1}= I_{2} - Insufficient data

Correct answer: 3. I_{1} = I_{2}

## Current Divider Rule MCQ#7

For the same circuit given in problem 6, if R_{1} > 10R_{2}, for any value of source current, the majority of current will flow through:

- R
_{1} - R
_{2}

Correct answer: 1. R_{2}

## Current Divider Rule MCQ#8

Current divider rule is never applied to series circuits because:

- It requires application of very complex formulae
- Current never divides in series

Correct answer: 2. Current never divides in series

## Current Divider Rule MCQ#9

For current divider circuit in problem 6, If I_{s} = 2 A and R_{1} = 3 Ω and R_{2} = 2 Ω and I_{1} and I_{2} are values of currents flowing through R_{1} and R_{2}, then:

- I
_{1}= 1.2 Ω, I_{2}= 0.8 Ω - I
_{1}= 1.5 Ω, I_{2}= 0.5 Ω - I
_{1}= 1.3 Ω, I_{2}= 0.7 Ω - I
_{1}= 0.8 Ω, I_{2}= 1.2 Ω

Correct answer: 4. I_{1} = 0.8 Ω, I_{2} = 1.2 Ω

## Current Divider MCQ#10

For circuit given in problem 6, if another resistor is added in series to R_{2}, the current flowing through R_{1} will:

- Increase
- Decrease
- Remain same

Correct answer: 1. Increase

By increasing resistance there will be reduction of current in R2. To understand this Consider initially current source is of 5A and resistor R1 is of 2 ohms, while R2 is also 2 ohms.

Now current across R1 is: 2.5 A; (I_{1} = (R_{eq}/R_{1}) * I_{t} = (1/2) * 5 A = 2.5 A

Now suppose another resistor (having resistance 1 ohms is added in series to R2. Now current across R1:

I_{1} = (R_{eq}/R_{1}) * I_{t} = (1.2/2) * 5 A = 3 A